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I am wondering if somebody has an example of use of Galmarino's test. The Galmarino test says that for $X=(X_t)_{t\in T}$ a continuous stochastic process with $\mathcal{F}$ the natural filtration, a random time $\tau$ is a stopping time if and only if for all $t\in T$, $\omega, \omega'\in\Omega$ $$\tau(\omega)\leq t, X_s(\omega)=X_s(\omega') \text{ for all }s\leq t \Rightarrow \tau(\omega')\leq t.$$

I can't find any examples. I did find a proof (Proving Galmarino's Test) and the answer to this question (Hitting time of an open set is not a stopping time for Brownian Motion) did help me to understand it a bit more. I tried to work out the case for canonical Brownian motion with $\tau=\inf\{t>0 \mid B_t\in (a,b)\}$ which is not a stopping time, as described in the answer. I understand why this $\tau$ is not a stopping time, but I can't wrap my head around why the condition is not satisfied.

I want to understand this test so I can use it to verify whether a random time is a stopping time. I hope somebody can help me.

Rinus94
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    Denote by $\omega$ and $\omega'$ the continuous functions from this picture, say $\omega$ is the one on the right and $\omega'$ is the one on the left. You can easily check that the condition from Galmarino's test fails for these functions (which are sample paths of the canonical Brownian motion). The point is essentially the following: Given a sample path $(X_s(\omega))_{s \leq t}$ up to time $t$, you cannot decide whether $\tau(\omega)=t$. – saz May 30 '18 at 09:55
  • Thanks! I get it now. I needed to get back to the basics of the infimum. The picture with your extra comment really helped me. – Rinus94 May 30 '18 at 11:09

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